# Halogenoalkanes

> CIE A-Level Chemistry · Unit 14: Halogen derivatives and alcohols
> Source: https://www.owlsprep.com/study/cie-9701-u14-halogenoalkanes/

Halogenoalkanes are key organic intermediates, formed by substituting alkane hydrogen with halogen atoms. This sub-topic covers their classification, physical properties, and core reaction types: nucleophilic substitution and elimination.

**Prerequisites:** [Alkane structure and nomenclature](https://www.owlsprep.com/study/cie-9701-u10-alkanes/); [Electronegativity and intermolecular forces](https://www.owlsprep.com/study/cie-9701-u03-chemical-bonding/); [Organic reaction mechanism basics](https://www.owlsprep.com/study/cie-9701-u11-intro-organic-mechanisms/)

## Learning objectives

- Classify halogenoalkanes as primary, secondary, tertiary based on structure
- Explain trends in physical properties of halogenoalkanes
- Distinguish between SN1 and SN2 nucleophilic substitution mechanisms
- Compare substitution and elimination reaction conditions and products

## Classification and Nomenclature

Halogenoalkanes (also called alkyl halides) are alkanes with at least one hydrogen atom replaced by a halogen atom. They are classified based on how many alkyl groups are bonded to the carbon that holds the halogen.

**Classification of Halogenoalkanes** — Classification is determined solely by the number of alkyl groups bonded to the halogen-bearing carbon atom, not the total number of carbons in the molecule.

*Example:* 1-bromopropane (1°), 2-bromopropane (2°), 2-bromo-2-methylpropane (3°)

**Worked example:** Classify 1-chloro-2-methylpropane and 2-chloro-2-methylbutane as primary, secondary or tertiary.

1. Step 1: Identify the carbon directly bonded to chlorine in 1-chloro-2-methylpropane. This terminal carbon is only bonded to one other alkyl carbon.
2. Thus, 1-chloro-2-methylpropane is a primary (1°) halogenoalkane.
3. Step 2: Identify the carbon directly bonded to chlorine in 2-chloro-2-methylbutane. This central carbon is bonded to three other alkyl carbons.
4. Thus, 2-chloro-2-methylbutane is a tertiary (3°) halogenoalkane.

> **Exam tip:** Always classify based on the halogen-bearing carbon, not the most substituted carbon elsewhere in the molecule.

## Physical Properties

The C-X bond is polar because halogens are more electronegative than carbon. However, halogenoalkanes cannot form hydrogen bonds with water, so they are immiscible with water. Boiling point depends primarily on molecular size.

| Compound | Relative Molecular Mass | Boiling Point (°C) |
| --- | --- | --- |
| Chloromethane | 50.5 | -24 |
| Bromomethane | 95 | 4 |
| Iodomethane | 142 | 43 |
| 1-chloropropane | 78.5 | 47 |

**Worked example:** Explain why 1-iodobutane has a higher boiling point than 1-chlorobutane.

1. Step 1: Iodine has a higher atomic number than chlorine, so 1-iodobutane has a higher relative molecular mass than 1-chlorobutane.
2. Step 2: Higher molecular mass leads to stronger instantaneous dipole-induced dipole (London) intermolecular forces between molecules.
3. Step 3: More thermal energy is required to overcome these stronger forces, resulting in a higher boiling point.

## Nucleophilic Substitution Mechanisms

Nucleophilic substitution is the most important reaction of halogenoalkanes. The electronegative halogen pulls electron density away from the carbon, making it electrophilic and open to attack by electron-rich nucleophiles.

**Nucleophilic Substitution** — A reaction where a nucleophile replaces the halide leaving group, donating a lone pair of electrons to the electrophilic carbon to form a new covalent bond.

Two mechanisms exist: SN2 (bimolecular, 1 step) for primary halogenoalkanes, and SN1 (unimolecular, 2 steps) for tertiary halogenoalkanes, via a stable carbocation intermediate.

**Worked example:** Describe the mechanism for the reaction of bromoethane with aqueous sodium hydroxide.

1. Step 1: Identify the nucleophile: hydroxide ion ($\text{OH}^-$) from NaOH. The C-Br bond is polar, so the carbon is $\delta^+$ and attracts the negatively charged nucleophile.
2. Bromoethane is a primary halogenoalkane, so it reacts via an SN2 mechanism (one step):
3. $$\ce{OH^- + CH3CH2Br -> HOCH2CH3 + Br^-}$$
4. Step 2: To draw the mechanism: draw a curly arrow from a lone pair on the $\text{OH}^-$ to the $\delta^+$ carbon, and a second curly arrow from the C-Br bonding pair to the bromine atom, showing the leaving group departing.

**Check your understanding**

Test your understanding:

1. Which of the following undergoes SN1 nucleophilic substitution fastest?

   - 1-bromobutane
   - 2-bromobutane
   - 2-bromo-2-methylpropane
   - bromomethane

   *Why:* Tertiary halogenoalkanes form stable tertiary carbocation intermediates, so they favour SN1 substitution over SN2.

> **Exam tip:** CIE examiners require curly arrows to start at a lone pair or bonding pair, not at the negative charge on the nucleophile.

## Elimination Reactions

When halogenoalkanes react with hot ethanolic (not aqueous) potassium hydroxide, elimination (also called dehydrohalogenation) occurs instead of substitution, forming an alkene by removing HX.

**Dehydrohalogenation (Elimination)** — A reaction that removes a hydrogen halide (HX) molecule from a halogenoalkane to form a carbon-carbon double bond (alkene).

*Example:* 2-bromopropane + hot ethanolic KOH → propene + KBr + H₂O

> **tip**
>
> The solvent is the key clue: aqueous hydroxide = substitution, hot ethanolic hydroxide = elimination.

**Worked example:** State the major organic product formed when 2-bromobutane reacts with hot ethanolic KOH.

1. Step 1: Elimination of HBr from 2-bromobutane can form two alkene products: but-1-ene and but-2-ene.
2. Step 2: Zaitsev's rule states the more substituted alkene (more alkyl groups attached to the double bond) is the major product.
3. Step 3: But-2-ene has two alkyl groups attached to the double bond, while but-1-ene only has one. Thus, but-2-ene is the major product.

## Common pitfalls

- **Wrong:** Classifying 1-chloro-2-methylpropane as a secondary halogenoalkane
  - Why it fails: Mistakenly counts the branched carbon instead of the carbon directly bonded to the halogen
  - Correct: Only classify based on the carbon directly attached to the halogen: 1-chloro-2-methylpropane is primary
- **Wrong:** Explaining higher boiling points of heavier halogenoalkanes by increasing polarity
  - Why it fails: Polarity decreases from chlorine to iodine, but boiling point increases, so polarity is not the main factor
  - Correct: Attribute boiling point trends to increasing relative molecular mass and stronger London intermolecular forces
- **Wrong:** Drawing a curly arrow from the negative charge of a nucleophile to the electrophilic carbon
  - Why it fails: Curly arrows represent the movement of an electron pair, not the negative charge
  - Correct: Always start the curly arrow at a lone pair on the nucleophile atom
- **Wrong:** Claiming primary halogenoalkanes undergo SN1 substitution
  - Why it fails: Primary carbocations are too unstable to form as intermediates
  - Correct: Primary halogenoalkanes always react via SN2 (1-step) substitution, while tertiary react via SN1
- **Wrong:** Using aqueous potassium hydroxide to form an alkene from a halogenoalkane
  - Why it fails: Aqueous conditions favour nucleophilic substitution to form an alcohol, not elimination
  - Correct: Use hot, ethanolic potassium hydroxide to carry out elimination and form an alkene

## Cheatsheet

| Property | Primary (1°) | Secondary (2°) | Tertiary (3°) |
| --- | --- | --- | --- |
| Halogen-bearing C bonded to | 1 other C | 2 other C | 3 other C |
| Substitution mechanism | SN2 (1 step) | Mixed SN1/SN2 | SN1 (2 step) |
| Intermediate formed | None | Variable | Carbocation |
| Product with aq NaOH | Alcohol | Alcohol | Alcohol |
| Product with hot ethanolic KOH | Alkene | Alkene | Alkene |

## What's next

Halogenoalkanes are fundamental building blocks for organic synthesis, used to introduce key functional groups like alcohols, amines, and alkenes that form the basis of more complex organic molecules. Mastery of their mechanisms is critical for CIE A-Level Chemistry, as mechanism questions frequently appear in both Paper 2 and Paper 4, requiring accurate drawing of curly arrows and identification of intermediates. Understanding the difference between substitution and elimination here will help you predict products of organic reactions across all subsequent topics, from synthesis to aromatic chemistry.

- [Alcohols](https://www.owlsprep.com/study/cie-9701-u14-alcohols/)
- [Carbonyl compounds and carboxylic acids](https://www.owlsprep.com/study/cie-9701-u15-overview/)
- [Aldehydes and Ketones](https://www.owlsprep.com/study/cie-9701-u15-aldehydes-and-ketones/)

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